Question 62

If $$x_0 = 1, x_1 = 2$$, and $$x_{n + 2} = \frac{1 + x_{n + 1}}{x_n}, n = 0, 1, 2, 3, ......,$$ then $$x_{2021}$$ is equal to

Solution

$$x_0=1$$

$$x_1=2$$

$$x_2=\frac{\left(1+x_1\right)}{x_0}=\frac{\left(1+2\right)}{1}=3$$

$$x_3=\frac{\left(1+x_2\right)}{x_1}=\frac{\left(1+3\right)}{2}=2$$

$$x_4=\frac{\left(1+x_3\right)}{x_2}=\frac{\left(1+2\right)}{3}=1$$

$$x_5=\frac{\left(1+x_4\right)}{x_3}=\frac{\left(1+1\right)}{2}=1$$

$$x_6=\frac{\left(1+x_5\right)}{x_4}=\frac{\left(1+1\right)}{1}=2$$

Hence, the series begins to repeat itself after every 5 terms. Terms whose number is of the form 5n are 1, 5n+1 are 2... and so on, where n=0,1,2,3,....

2021 is of the form 5n+1. Hence, its value will be 2.

Video Solution

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